---
title: "Derivation of Betz limit for wind turbine"
subject: "Energy Resources"
chapter: 4
key: "semester_6/energy_resources:betz-limit-derivation-wind"
source: https://www.bctengineer.com/q/semester_6/energy_resources/betz-limit-derivation-wind
site: BCT Engineering
---

# Derivation of Betz limit for wind turbine

Energy Resources · Chapter 4: Wind energy

- Asked in: year not given, 8 marks
- Video: [One Dimensional Momentum Theory and the Betz Limit for Maximum Power Production](https://www.youtube.com/watch?v=HJooZQUXr6o) — DTU Wind and Energy Systems
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/energy_resources/betz-limit-derivation-wind
- Whole chapter: https://www.bctengineer.com/md/semester_6/energy_resources/chapter-4

## Question

Derive an expression for the power extracted by an ideal wind turbine using actuator disc theory and show that the maximum power coefficient is 16/27 (Betz limit). What does this value mean in practice?

## Short answer

Using actuator disc theory and momentum conservation, power coefficient Cp = ½(1+a)(1-a²) with a = v2/v1. The maximum occurs at a = 1/3, giving Cp,max = 16/27 ≈ 0.593. This Betz limit means no wind turbine can extract more than 59.3% of the wind's kinetic energy; practical turbines achieve 0.40-0.50 due to drag and tip losses.

## Answer

#### Assumptions
Ideal rotor as an actuator disc, steady, incompressible, frictionless flow, uniform velocity over the disc and no rotation of the wake. Let $v_1$ be the upstream (free) wind speed, $v$ the speed through the disc of area $A$, and $v_2$ the speed far downstream.

```text
  v1 ->  \   |  /  -> v2 (slow)
 --------\--|--/--------
 ->      | disc |  v
 --------/--|--\--------
```

#### Power from momentum
Mass flow through the disc: $\dot m = \rho A v$.

Force on the disc (change of momentum): $F = \dot m (v_1 - v_2)$.

Power extracted: $P = Fv = \rho A v^2 (v_1 - v_2)$.

Power also equals the loss of kinetic energy of the stream:

$$
P = \tfrac{1}{2}\dot m (v_1^2 - v_2^2) = \tfrac{1}{2}\rho A v (v_1 - v_2)(v_1 + v_2)
$$

Equating the two expressions for $P$:

$$
v = \frac{v_1 + v_2}{2}
$$

So the wind speed at the disc is the mean of the upstream and downstream speeds.

#### Power coefficient
Substituting $v$:

$$
P = \tfrac{1}{4}\rho A (v_1 + v_2)(v_1^2 - v_2^2)
$$

Let $a = v_2/v_1$. Then

$$
P = \tfrac{1}{2}\rho A v_1^3 \cdot \tfrac{1}{2}(1+a)(1-a^2)
$$

The available power in the wind is $P_0 = \tfrac{1}{2}\rho A v_1^3$, so

$$
C_p = \frac{P}{P_0} = \tfrac{1}{2}(1+a)(1-a^2)
$$

#### Maximum
$$
\frac{dC_p}{da} = \tfrac{1}{2}\left(1 - 2a - 3a^2\right) = 0 \Rightarrow 3a^2 + 2a - 1 = 0
$$

giving $a = \tfrac{1}{3}$ (the other root is $-1$, not physical). Then

$$
C_{p,max} = \tfrac{1}{2}\left(\tfrac{4}{3}\right)\left(\tfrac{8}{9}\right) = \frac{16}{27} = 0.593
$$

#### Meaning
No wind turbine can capture more than 59.3 % of the kinetic energy of the wind passing through its swept area, because the air must keep moving behind the rotor. Practical large turbines reach $C_p$ of 0.40–0.50 due to blade drag, tip losses and wake rotation; the maximum for a Savonius rotor is about 0.15–0.20.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the wind speed through the disc?
   A. v = (v1 + v2)/2
   B. v = v1
   C. v = v1 x v2
   D. v = v1 - v2
   Answer: A. Equating momentum power ρAv²(v1-v2) with kinetic energy loss gives the mean. (Key point: Disc velocity)

2. With a = v2/v1, Cp equals...
   A. (1 - a)²
   B. a(1 - a)
   C. (1 + a)²(1 - a)
   D. ½(1 + a)(1 - a²)
   Answer: D. P = ½ρAv1³ x ½(1+a)(1-a²). (Key point: Power coefficient)

3. At what a is Cp maximum, and what is Cp?
   A. a = 1, Cp = 1
   B. a = 1/2, Cp = 0.5
   C. a = 1/3, Cp = 16/27
   D. a = 1/3, Cp = 8/27
   Answer: C. Solving 3a² + 2a - 1 = 0 gives a = 1/3, and ½(4/3)(8/9) = 16/27. (Key point: Optimum ratio)

4. What does the Betz limit mean in practice?
   A. All turbines reach 59.3%
   B. No turbine can capture more than 59.3% of wind kinetic energy because the air must keep moving
   C. Turbines can reach 100% if frictionless
   D. It is the minimum efficiency of a rotor
   Answer: B. Real turbines reach Cp 0.40-0.50. (Key point: Practical meaning)

## Related in Energy Resources

- [Wind turbine power, tip speed ratio and energy](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Awind-turbine-power-energy-numerical) — Energy Resources, chapter 4
- [Horizontal and vertical axis wind turbines](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Ahawt-vawt-types-characteristics) — Energy Resources, chapter 4

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